Engineering Calculus Notes 295

Engineering Calculus Notes 295 - − 3 2 through y = 1 2 ,...

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3.5. SURFACES AND THEIR TANGENT PLANES 283 -1 0 1 2 -3 -2 -1 0 1 b −→ v y Figure 3.11: Slicing the graph of x 2 3 y 2 2 at x = 1 which is the graph of the function z = 1 2 3 y 2 2 ; at y = 1 2 , the derivative of this function is dz dy v v v v 1 2 = ∂f ∂y p 1 , 1 2 P = 3 2 and −→ v 2 = −→ 3 2 −→ k is the direction vector for the line of slope
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Unformatted text preview: − 3 2 through y = 1 2 , z = 1 8 in this plane—a line which also lies in the tangent plane. This line is parametrized by x = 1 y = 1 2 + t z = 1 8 − 3 2 t...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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